有没有办法在 C++ 的赋值中将“char**”转换为“double”？

Question

我试图将用户输入的执行限制为仅数字，当他们输入 a / b 而不是 8 / 2 时，结果为零，但是当他们尝试输入这样的内容时，我想给他们一个警告。当我执行我的代码时，我收到无法转换char**为double赋值的错误。我为转换实现的字符是numvalidation验证用户输入的第一个值，如果输入是数字，则接受否则返回默认消息。

#include <iostream>
using namespace std;
int main()
{
    char * numvalidation;
    double var1, var2;
    beginning:
    cout << "Enter the First value: ";
    cin >> var1;
    cout << "Enter the second value: ";
    cin >> var2;
    if (var1 != '\0')
    {
        var1 = (&numvalidation);
        if (*numvalidation != '\0')
        {
            cout << "First Value was not a number,Try again with a correct value." << endl;
        }
        {
            cout << "Do you want to continue using the Calculator? (Y/N)" << endl;
            char restart;
            cin >> restart;
            if (restart == 'y' || restart == 'Y')
                goto beginning;
        }
    }
    cout << "What do you want to do?" << endl;
    cout << "+ :Addition is available" <<endl;
    cout << "- :Subtraction is available" <<endl;
    cout << "* :Multiplication is available" <<endl;
    cout << "/ :Division is available" <<endl;
    cout << "Please chose one of the option below" <<endl;
    char decision;
    cout << "Decision: ";
    cin >> decision;
    system ("cls");
    switch (decision)
    {
        case '+':
            cout << var1 << " + " << var2 << " = " << "" << ( var1 + var2 ) <<endl;
            break;
        case '-':
            cout << var1 << " - " << var2 << " = " << "" << ( var1 - var2 ) <<endl;
            break;
        case '*':
            cout << var1 << " * " << var2 << " = " << "" << ( var1 * var2 ) <<endl;
            break;
        case '/':
            if (var2 == !0)
                cout << var1 << " / " << var2 << " = " << "" << ( var1 / var2 ) <<endl;
            else
                cout << "The value you have entered is invalid because you cannot divide any number by zero " <<endl;
            break;
        default:
            cout << "You have not entered the correct data";
    }
    {
        cout << "Do you want to continue using the Calculator? (Y/N)" << endl;
        char decision2;
        cin >> decision2;
        if (decision2 == 'y' || decision2 == 'Y')
            goto beginning;
    }
}

Answer 1

下面的代码要求一个双精度直到它得到一个有效的双精度。

它使用std::string. 您只能使用，isdigit()但这会很荒谬（在我看来）。

#include<string>
#include<iostream>
#include<cctype>

int main(){

    double var1{};
    std::string temp{};
    size_t processed_chars_no{};


    do{
        std::cout<<"Enter a valid double value: \n";
        std::cin >> temp;
        if( isdigit(temp[0]) )var1 = std::stod( temp, &processed_chars_no );
    }
    while ( processed_chars_no != temp.size() );

    std::cout<<"\n"<<var1;

}